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Stream: learning: questions

Topic: Help with a false proof about pullbacks and monomorphisms

Olli (Dec 21 2020 at 09:22):

I came up with a false proof that says that maps from the pullback object to the target object of the pullback are always monic, here it is:

1. Suppose $fh = gk$ is a pullback square, and that $f(hy) = g(kx)$ is a commuting square.

2. The universal property of pullbacks says that the maps $hy$ and $kx$ have unique factorizations, which amounts to saying that $x = y$.

3. Let's call the map $fh = gk$ from the pullback object to the pullback target as $p$. So then what I have just stated is that from $px = py$ we can conclude that $x = y$, so this means that $p$ is monic.

But this can't be true as it would imply for example that in $Set$ the map $A \times B \to 1$ is always monic. Where is the mistake?

Olli (Dec 21 2020 at 09:22):

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Tom de Jong (Dec 21 2020 at 09:43):

Olli said:

I came up with a false proof that says that maps from the pullback object to the target object of the pullback are always monic, here it is:

1. Suppose $fh = gk$ is a pullback square, and that $f(hy) = g(kx)$ is a commuting square.

2. The universal property of pullbacks says that the maps $hy$ and $kx$ have unique factorizations, which amounts to saying that $x = y$.

3. Let's call the map $fh = gk$ from the pullback object to the pullback target as $p$. So then what I have just stated is that from $px = py$ we can conclude that $x = y$, so this means that $p$ is monic.

But this can't be true as it would imply for example that in $Set$ the map $A \times B \to 1$ is always monic. Where is the mistake?

I think that item 2 is false, since the universal property of the pullbacks only gives you an arrow $z$ unique with the property that $kz = kx$ and $hz = hy$, but this doesn't necessarily imply $x=y$ (unless $k$ and $h$ happen to be monic, for instance).

Morgan Rogers (he/him) (Dec 21 2020 at 10:36):

It should be noted that a variation on this 'proof' allows you to prove the true statement that the pair $(h,k)$ is jointly monic in a pullback square like the one you described.

Javier Prieto (Dec 21 2020 at 13:13):

For $p$ to be monic, you'd need $x$ and $y$ to range over all morphisms of the right type independently. This is not granted by the universal property of the pullback

Olli (Dec 21 2020 at 17:26):

Thanks, that does make sense and clears up my confusion.